# When Does $\sigma(q^k)$ Have a Prime Factor Greater Than $q$

Let $$q$$ be prime and $$k$$ be a natural number. When does $$\sigma(q^k)$$ have a prime factor greater than $$q$$?

We can slightly reduce the problem by noting that $$\sigma(q^k)=\frac{q^{k+1}-1}{q-1}$$ Thus if $$p\nmid q-1$$, then $$p\mid \sigma(q^k)$$ if and only if $$p\mid q^{k+1}-1$$. Clearly a prime $$p>q$$ does not divide $$q-1$$, so $$\sigma(q^k)$$ has a prime factor greater than $$q$$ if and only if $$q^{k+1}-1$$ does.

Using this restatement, if one takes a look at one of my answers to this question we see this is the case when $$k+1\ge q$$ letting $$a=q$$, $$b=1$$, and $$n=k+1$$ (note the question requires $$b>1$$, but the proof only uses Zsigmondy's Theorem in which $$b>0$$ suffices). However clearly this is true in more than just this case. Can we describe all cases in which this is true, or at least expand the set of $$(q,k)$$ for which we know that this holds?

Edit
Lets define $$\kappa(n)$$ to be the smallest prime congruent to $$1\mod n$$. A modified argument of that in the linked question, where we restrict $$b$$ to $$1$$, leads one to conclude that there exists a prime $$\ge \kappa(n)$$ that divides $$q^n-1$$. Thus if $$\kappa(k+1)\ge q$$ (or really $$\kappa(k+1)\ge q-1$$ since $$q\nmid q^n-1$$), then $$\sigma(q^k)$$ satisfies the problem, slightly increasing the set of valid $$(q,k)$$.

• When $k+1$ is even, Because number of terms of series representing sum of divisors is equal to $k+1$ and if it is even then there is a common factor like $q+1$ in it; $\sigma=[q^k(q+1)+q^{k-1}(q+1)+ . . . +(q+1) ]$. Feb 18, 2019 at 12:21
• @sirous However, unless $q=2$, since $q$ is a prime $2\mid q+1$ and so any prime factors contributed by $q+1$ will be smaller than $q$. Feb 18, 2019 at 17:42
• the only even prime i.e. 2 always is disturbing. What I claim is for odd primes and is correct. Feb 19, 2019 at 6:31
• Wont you down vote me if I post this as an answer? Feb 19, 2019 at 11:53
• @sirous the thing is you found a factor but I’m looking for prime factors. Feb 19, 2019 at 12:53